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all x e E, then/= sup/n with/n = hn/(l - hn\ and/w is finite and continuous


on E.

So we are now in the situation where 0 rg/:g 1 and/has compact support.
For each integer n ^ 1 , consider the. decreasing finite sequence of open sets
AfaI=:/~1Qfc/H, +00]) (12.7.2), where O^fc^w-1. These sets Akn are

n- 1

relatively compact. The function #„ = (l/ri) ]T cpAkn is lower semicontinuous


(12.7.5), and we have 0 ^f(x) - gn(x) g 1/fl for all x e E, hence/is the upper
envelope of the sequence (#„). On the other hand, since each cpAkn is the upper
envelope of a sequence (//jtm«)m£i of continuous functions, it follows that#n is

n- 1

the upper envelope of the sequence (hmn)m^ where hmn = OAO E **»•». (2.3.11),


and the hmn are continuous. Since gn ^ 1 — (!/«)> tne functions Amn do not take
the value 1 ? and it is clear that/is the upper envelope of the " double " family
(/uUi.^i (2-3.6).

(12.7.9) Let "E be a nonempty metrizable compact space and let f be a lower
semicontinuous mapping ofE into R. Then there exists at least one point aeE
such that f (a) = inf/(x) (in other words, f attains its lower bound on E).


Let ju = inf f(x). Then there exists a decreasing sequence (AM) of elements


of R, belonging to/(E) and such that inf /ln = ju. The set/"1^) is therefore the


intersection of the nonempty closed sets Fn =/~1([~~00? ^-J) (12.7.2), which
form a decreasing sequence. If their intersection were empty, the open sets
Un = E — Fn would cover E. But E is compact and the sequence (Un) is
increasing; hence we should have one of the Un equal to E, which is absurd.

In particular, if / does not take the value — oo on E, then / is minorized

Let (xn) be any sequence of elements of R, and put
j;n=infxB+p,       zn

The sequence (yn) (resp. (zn)) is increasing (resp. decreasing), hence has a
limit (4,2.1). We write

(12.7.10)   lim inf xn =lim/inf xn+p\,       lim sup xn =lim /sup xn+p\

«-*-oo                  M-*OO \JpisO             /                   W-+00                  n-*cxD Vp^O              /

Clearly we have lim sup xn = — lim inf( — xn), so that we need only consider

n-*oo                          /i-*oo

the limit inferior.